InterviewSolution
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InterviewSolution
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Sounds from two identical sources `S_(1)` and `S_(2)` reach a point `P`. When the sounds reach directly, and in the same phase, the intensity at `P` is `I_(0)`. The power of `S_(1)` is now reduced by `64%` and the phase difference between `S_(1)` and `S_(2)` is varied continuously. The maximum and minimum intensities recorded at `P` are now `I_("max")` and `I_("min")`A. `I_(max) = 0.64 I_(0)`B. `I_(min) = 0.36 I_(0)`C. `(I_(max))/(I_(min)) = 16`D. `(I_(max))/(I_(min)) = (1.64)/(0.36)` |
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Answer» Correct Answer - A::C Let a= Initial amplitude due to `S_(1)` and `S_(2)` each `I_(0) = k(4a)^(2)`, where k is a constant After reduction of power of `S_(1)`, Due to superposition `a_(max)=a+0.6a=1.6a,` and `a_(min)=1-0.6a=0.4a` `I_(max)//I_(min)=(a_(max)//a_(min))^(2)=(1.6a//0.4a)=16` |
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